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09 07 14

Esame completo di Advanced Computer Architectures per il corso di Computer Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

Advanced Computer ArchitecturesEsame completo

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Esame completo di Advanced Computer Architectures per il corso di Computer Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

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Course on Advanced Computer Architectures Surname (Cognome) SOLUTION Name (Nome) POLIMI ID Number POLIMI Email Address Signature (Firma) Politecnico di Milano, July 9th, 2014 Prof. C. Silvano EX1A ( 2.5points) EX1B ( 2.5points) EX1C ( 2.5points) EX2A ( 5 points) EX2B ( 2.5 points) Q3 ( 5 points) Q4 ( 5 points) Q5 ( 5 points) Q6 ( 3 points) TOTAL (33 points) (33 points) Course on: Advanced Computer Architectures – Prof. C Silvano EXAM 09/07/2014 – Please write in CAPITAL LETTERS AND BLACK/BLUE COLORS!!! (MAIUSCOLO e COLORE NERO/BLU!!!) Page 1 - SOLUTION EXERCISE 1A – PIPELINE BASIC (2.5 points) Given the following loop expressed in a high level language: for (i =0; i < N; i ++)  vectB[i] = vectB[i] + k1 + k2;  The program has been compiled in MIPS assembly code assuming that registers $t6 and $t7 have been initialized with values 0 and 4N respectively. The symbol VECTB is a 16-bit constant. The processor clock frequency is 1.2 GHz. Let us consider the loop executed by 5-stage pipelined MIPS processor WITHOUT any optimization in the pipeline (PLEASE don’t consider any inter-iteration dependencies) 1. Identify the RAW (Read After Write) Hazards in the pipeline scheme and identify the Hazard Type (Data Hazard or Control Hazard) in the last column 2. Identify in the first column the number of stalls to be inserted before each instruction (or between the stage IF and ID of each instruction) necessary to solve the hazards Num. Stalls INSTRUCTION C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 Hazard Type 3 FOR:beq $t6,$t7,END IF ID EX ME WB CNTR 3 lw $t2,VECTB($t6) IF ID EX ME WB CNTR 3 addi $t2,$t2,k1 IF ID EX ME WB RAW $t2 3 addi $t2,$t2,k2 I F ID EX ME WB RAW $t2 3 sw $t2,VECTB($t6) IF ID EX ME WB RAW $t2 addi $t6,$t6,4 IF ID EX ME WB 3 blt $t6,$t7, FOR IF ID EX ME WB RAW $t6 blt…

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